Question

The diameter of the lower and upper ends of a bucket (in the form of a frustum of cone) are 10cm and 30cm respectively.if its height is 24cm, find the area of the metal sheet used to make the bucket

Answer 1

Solution :

_____________________________________________________________

Given :

Statement :
            The diameter of the lower and upper ends of a bucket (in the form of a frustum of cone) are 10cm and 30cm respectively,

We know that,

⇒ Radius is half of diameter

Hence,

⇒ $r = \frac{d}{2}$

⇒ $r_1 = \frac{30}{2} = 15$ cm, (Radius of upper end),.

⇒ ∴$r_1 = 15$cm

⇒ $r_2 = \frac{10}{2} = 5$ cm, (Radius of lower end),.

⇒ ∴ $r_2 = 5$cm,.

Height of the bucket (Frustum of the cone) = 24cm

Hence it's slant height = l = $\sqrt{h^2 + (r_1 - r_2)^2}$

⇒ $\sqrt{24^2 + (15 - 5)^2}$

⇒ $\sqrt{24^2 + (10)^2}$

⇒ $\sqrt{576 + 100}$

⇒ $\sqrt{676}$

⇒ l = 26 cm

       ∴ The slant-height of the frustum = 26 cm.

_____________________________________________________________

To Find :

The area of metal used to make the bucket,.

_____________________________________________________________

We know that,

Bucket has the dimensions of:

(i) CSA of frustum,

(ii) Area of bottom (as bucket's area of bottom)

_______________________________________

Hence, we need to find :

Area of metal used = The CSA of bucket + Area of base

⇒ $\pi l(r_1 + r_2) + \pi r_2^2$

⇒ $\pi [ l(r_1 + r_2) +  r_2^2]$

Taking $\pi = \frac{22}{7}$ we get,

⇒ $\frac{22}{7}[(26)(15 + 5) +(5)^2]$

⇒ $\frac{22}{7}[(26)(20) + 25]$

⇒ $\frac{22}{7}(520 + 25)$

⇒ $\frac{22}{7}(545)$

⇒ $\frac{11990}{7}$

⇒ 1712.86 cm².

  ∴ The area of metal used = 1712.86 cm²

_____________________________________________________________

                                        Hope it Helps!!